What is the integral of 1/(2x+x^3) ?

The integral of 1/(2x+x^3) is 1/2 ln|x|-1/4 ln|x^2+2|+C

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integral of 1/(2x+x^3)

\int \frac{1}{2 x + x ^{3}} d x

\frac{1}{2} ln ∣ x ∣ - \frac{1}{4} ln x^{2} + 2 + C

Solution steps

\int \frac{1}{2 x + x ^{3}} d x

Take the partial fraction of

\frac{1}{2 x + x ^{3}} : \frac{1}{2 x} - \frac{x}{2 ( x ^{2} + 2 )}

= \int \frac{1}{2 x} - \frac{x}{2 ( x ^{2} + 2 )} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{1}{2 x} d x - \int \frac{x}{2 ( x ^{2} + 2 )} d x

\int \frac{1}{2 x} d x = \frac{1}{2} ln ∣ x ∣

\int \frac{x}{2 ( x ^{2} + 2 )} d x = \frac{1}{4} ln x^{2} + 2

= \frac{1}{2} ln ∣ x ∣ - \frac{1}{4} ln x^{2} + 2

Add a constant to the solution

= \frac{1}{2} ln ∣ x ∣ - \frac{1}{4} ln x^{2} + 2 + C

\int \frac{1}{x 64 x ^{4} - 1} d x

\frac{d}{d x} (\frac{1}{( x ^{3} - 2 x ^{2} + 7 ) ^{4}})

n = 1 \sum \infty 4 n - 1

\int \frac{1}{1 + 3 x ^{2}} d x

What is the integral of 1/(2x+x^3) ?
The integral of 1/(2x+x^3) is 1/2 ln|x|-1/4 ln|x^2+2|+C